A New Description of the Complex Roots Using Zeros of Some Polynomials

Abstract

Inspired by a problem found in a textbook of Faddeev and Sominsky we propose a new form of roots of an arbitrary complex number. Our idea uses the zeros of some polynomials connected with the modified Chebyshev polynomials.

1 Results

In the known textbook [1] authored by Faddeev and Sominsky one can encounter the following task No 171.

Task 1

Show that

$$\begin{aligned} \frac{u(a+bi)+u^2i}{2a\root 3 \of {a^2+b^2}}\in \root 3 \of {a+bi}, \end{aligned}$$

where u is a zero of the equation $u^3-3(a^2+b^2)u-2b(a^2+b^2)=0$.

Unfortunately, the above claim does not hold, what can be observed on the following example.

Example 1

For $a=1$ and $b=1$ the equation defined in Task 1 is as follows:

$$\begin{aligned} u^3-6u-4=0. \end{aligned}$$

Its zeros are equal to: $-2$, $1+\sqrt{3}$, $1-\sqrt{3}$. Therefore, the inclusion

$$\begin{aligned} \left\{ -\frac{1}{\root 3 \of {2}}+\frac{1}{\root 3 \of {2}}i,\frac{1+\sqrt{3}}{2\root 3 \of {2}}+\frac{5+3 \sqrt{3}}{2 \root 3 \of {2}}i,\frac{1-\sqrt{3}}{2\root 3 \of {2}}+\frac{5-3 \sqrt{3}}{2 \root 3 \of {2}}i\right\} \subset \root 3 \of {1+i}, \end{aligned}$$

should hold. Yet, it does not, as we have the equality:

$$\begin{aligned} \left\{ -\frac{1-i}{\root 3 \of {2}},\frac{1+\sqrt{3}}{2 \root 3 \of {2}}+\frac{\sqrt{3}-1}{2\root 3 \of {2}}i,\frac{1-\sqrt{3}}{2 \root 3 \of {2}}-\frac{1+\sqrt{3}}{2 \root 3 \of {2}}i\right\} =\root 3 \of {1+i}. \end{aligned}$$

Nevertheless, the idea hidden in Task 1 can be somewhat executed, and even, generalized, what is the aim of the present work. The main theorem of the present article uses the modified Chebyshev polynomials of the first kind. Let us recall (see [2, 3]) that for every $n\in {\mathbb {N}}_0$ the Chebyshev polynomial $T_n(x)$ of the first kind is a polynomial of degree n defined by the relation

$$\begin{aligned} T_n(\cos \theta )=\cos n\theta , \qquad \theta \in [0,\pi ]. \end{aligned}$$

We have

$$\begin{aligned} T_0(x)&=1, \qquad T_1(x)=x, \qquad T_2(x)=2x^2-1,\\ T_3(x)&=4x^3-3x, \qquad T_4(x)=8x^4-8x^2+1. \end{aligned}$$

The polynomials ${\varOmega }_n(x):=2T_n\left( \frac{x}{2}\right)$ ($n\in {\mathbb {N}}_0$) are called modified Chebyshev polynomials and satisfy the reccurent relation

$$\begin{aligned} {\varOmega }_0(x)&=2, \qquad {\varOmega }_1(x)=x,\\ {\varOmega }_{n+2}(x)&=x{\varOmega }_{n+1}(x)-{\varOmega }_n(x) \end{aligned}$$

for every $n\in {\mathbb {N}}_0$.

Now let us introduce two new families of polynomials associated with ${\varOmega }_n(x)$. Let $a, b\in {\mathbb {R}}$ with $a^2+b^2 \ne 0$ and let us put $\lambda :=\sqrt{a^2+b^2}$.

Definition 1

We call the following complex polynomial:

$$\begin{aligned} V_n^{(a,b)}(x)={\varOmega }_n(x)-\frac{2a}{\lambda } \end{aligned}$$

the first polynomial associated with the modified Chebyshev polynomial, whereas the complex polynomial

$$\begin{aligned} W_n^{(a,b)}(x)={\varOmega }_n(x)-\frac{2b}{\lambda }\sin \frac{\pi n}{2} \end{aligned}$$

is called the second polynomial associated with the modified Chebyshev polynomial.

The polynomials defined above admit the following factorization.

Lemma 1

For $n\in 2{\mathbb {N}}+1$ we have

$$\begin{aligned} V_n^{(a,b)}(x)=\prod _{k=0}^{n-1} \left( x-2\cos \left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) \right) \end{aligned}$$

(1)

and

$$\begin{aligned} W_n^{(a,b)}(x)=\prod _{k=0}^{n-1} \left( x-2\sin \left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) \right) , \end{aligned}$$

(2)

where $\varphi \in [0,2\pi )$ is an angle such that

$$\begin{aligned} \cos \varphi =\frac{a}{\lambda }\qquad \text {and}\qquad \sin \varphi =\frac{b}{\lambda }. \end{aligned}$$

Proof

In [4] one can find the following identity that holds for any $t\in {\mathbb {R}}$ and $n\in 2{\mathbb {N}}+1$:

$$\begin{aligned} {\varOmega }_n(x)-2\cos (nt)=\prod _{k=0}^{n-1} \left( x-2\cos \left( t+\frac{2\pi k}{n}\right) \right) . \end{aligned}$$

(3)

To show that (1) holds, let us replace the angle t from (3) by $\varphi$ satisfying the relation $t=\frac{\varphi }{n}$:

$$\begin{aligned} {\varOmega }_n(x)-\frac{2a}{\sqrt{a^2+b_2}}={\varOmega }_n(x)-2\cos \varphi {\mathop {=}\limits ^{(3)}}\prod _{k=0}^{n-1} \left( x-2\cos \left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) \right) , \end{aligned}$$

what immediately proves (1). In order to show that (2) holds, let us now replace the angle t from (3) by $\varphi$ satisfying the relation $t=\frac{\pi }{2}-\frac{\varphi }{n}$. Using trigonometric identities and fact that $n\in 2{\mathbb {N}}+1$, we obtain

$$\begin{aligned}&{\varOmega }_n(x)-\frac{2b}{\lambda }\sin \frac{\pi n}{2}={\varOmega }_n(x)-2\sin \varphi \sin \frac{\pi n}{2}\\&\quad ={\varOmega }_n(x)-2\cos \left( \frac{\pi n}{2}-\varphi \right) {\mathop {=}\limits ^{(3)}}\prod _{k=0}^{n-1} \left( x-2\cos \left( \frac{\pi }{2}-\frac{\varphi }{n}+\frac{2\pi k}{n}\right) \right) \\&\quad =\prod _{k=0}^{n-1} \left( x-2\cos \left( \frac{\pi }{2}-\left( \frac{\varphi }{n}-\frac{2\pi k}{n}\right) \right) \right) =\prod _{k=0}^{n-1} \left( x-2\sin \left( \frac{\varphi }{n}-\frac{2\pi k}{n}\right) \right) \\&\quad =\prod _{k=0}^{n-1} \left( x-2\sin \left( \frac{\varphi }{n}+\frac{2\pi (n- k)}{n}\right) \right) . \end{aligned}$$

Using the substitution of indices $j=n-k$, we get

$$\begin{aligned} {\varOmega }_n(x)-\frac{2b}{\lambda }\sin \frac{\pi n}{2}= \prod _{j=1}^n \left( x-2\sin \left( \frac{\varphi }{n}+\frac{2\pi j}{n}\right) \right) . \end{aligned}$$

Moreover, from a trigonometric identity, it follows:

$$\begin{aligned} {\varOmega }_n(x)-\frac{2b}{\lambda }\sin \frac{\pi n}{2}= \prod _{j=0}^{n-1} \left( x+2\sin \left( \frac{\varphi }{n}-\frac{2\pi j}{n}\right) \right) . \end{aligned}$$

Clearly, latter yields (2). $\square$

The below lemma establishes the relation between the roots of polynomials $V_n^{(a,b)}$, $W_n^{(a,b)}$ and roots of the n-th degree of the number $a+ib$.

Lemma 2

Suppose $n\in 2{\mathbb {N}}+1$ and that v, w are zeros of the polynomials $V_n^{(a,b)}$, $W_n^{(a,b)}$, respectively. If the equality

$$\begin{aligned} v^2+w^2=4 \end{aligned}$$

holds, then

$$\begin{aligned} \frac{\root n \of {\lambda }(v+iw)}{2}\in \root n \of {a+ib}. \end{aligned}$$

Proof

Let $\varphi \in [0,2\pi )$ be an angle such that

$$\begin{aligned} \cos \varphi =\frac{a}{\lambda }\qquad \text {and}\qquad \sin \varphi =\frac{b}{\lambda }. \end{aligned}$$

If v, w are zeros of $V_n^{(a,b)}$, $W_n^{(a,b)}$, respectively, then by Lemma 1:

$$\begin{aligned} v=2\cos \left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) \qquad \text {and} \qquad w=2\sin \left( \frac{\varphi }{n}+\frac{2\pi j}{n}\right) , \end{aligned}$$

for some $k,j\in \{0, 1, \ldots , n-1\}$. Let u, v satisfy the assumption $v^2+w^2=4$. The latter is equivalent to

$$\begin{aligned} 4\cos ^2\left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) +4\sin ^2\left( \frac{\varphi }{n}+\frac{2\pi j}{n}\right) =4. \end{aligned}$$

Using the trigonometric identity we obtain that it is also equivalent to:

$$\begin{aligned} \cos ^2\left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) =\cos ^2\left( \frac{\varphi }{n}+\frac{2\pi j}{n}\right) . \end{aligned}$$

(4)

From (not only) trigonometric identities we get

$$\begin{aligned} 0&=\cos ^2\left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) -\cos ^2\left( \frac{\varphi }{n}+\frac{2\pi j}{n}\right) \\&=-\sin \left( \frac{2\varphi }{n}+\frac{2\pi (k+j)}{n}\right) \sin \frac{2\pi (k-j)}{n}. \end{aligned}$$

If $\varphi \in [0, 2\pi ){\setminus }\left\{ 0, \frac{1}{2}\pi , \pi , \frac{3}{2}\pi \right\}$, then one can check that

$$\begin{aligned} \sin \left( \frac{2\varphi }{n}+\frac{2\pi (k+j)}{n}\right) \ne 0. \end{aligned}$$

Thus, Equation (4) is satisfied if and only if

$$\begin{aligned} \sin \frac{2\pi (k-j)}{n}=0, \end{aligned}$$

so one of the following cases holds:

$$\begin{aligned} j=k-\frac{n}{2}, \quad j=k \quad \text {or} \quad j=k+\frac{n}{2}. \end{aligned}$$

However, as n is odd, only the case $j=k$ is possible. Thus, we have

$$\begin{aligned} v=2\cos \left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) \qquad \text {and} \qquad w=2\sin \left( \frac{\varphi }{n}+\frac{2\pi k}{n}\right) . \end{aligned}$$

Consider now the cases $\varphi \in \left\{ 0, \frac{1}{2}\pi , \pi , \frac{3}{2}\pi \right\}$. Let first $\varphi =0$. Then Equation (4) is satisfied if and only if

$$\begin{aligned} 0=-\sin \frac{2\pi (k+j)}{n}\sin \frac{2\pi (k-j)}{n}. \end{aligned}$$

When $\sin \frac{2\pi (k-j)}{n}$ we obtain (analogously as above) $j=k$ and therefore

$$\begin{aligned} v=2\cos \frac{2\pi k}{n} \qquad \text {and} \qquad w=2\sin \frac{2\pi k}{n}. \end{aligned}$$

Whereas in the case

$$\begin{aligned} \sin \frac{2\pi (k+j)}{n}=0, \end{aligned}$$

we have

$$\begin{aligned} k=-j, \quad k=\frac{n}{2}-j, \quad k=n-j \quad \text {or} \quad k=\frac{3n}{2}-j, \end{aligned}$$

where the equality $k=n-j$ is consistent. However, one has

$$\begin{aligned} v=2\cos \frac{2\pi (n-j)}{n}=2\cos \frac{2\pi j}{n}, \end{aligned}$$

so

$$\begin{aligned} w=2\sin \frac{2\pi j}{n}. \end{aligned}$$

The other three cases can be checked similarly. $\square$

Using Lemma 2 we can give a full description of the set $\root n \of {a+ib}$ for odd n that, in fact, is a generalization of the correct form of Task 1.

Theorem 1

Suppose $n\in 2{\mathbb {N}}+1$. If $v_k$ and $w_k$, where $k\in \{1, 2, \ldots , n\}$, are zeros of the polynomials $V_n^{(a,b)}$ i $W_n^{(a,b)}$, respectively, such that

$$\begin{aligned} |v_1|\le |v_2|\le \ldots \le |v_n| \qquad \text {and} \qquad |w_1|\ge |w_2|\ge \ldots \ge |w_n|, \end{aligned}$$

then the following equality holds:

$$\begin{aligned} \left\{ \frac{\root n \of {\lambda }(v_k+iw_k)}{2}:k\in \{1, 2, \ldots , n\}\right\} =\root n \of {a+bi}. \end{aligned}$$